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Unproven claim in Weinberg's textbook

  1. Sep 10, 2013 #1
    In his Lectures on Quantum Mechanics, Weinberg makes the following claim about solutions to the Schrodinger equation for a central potential: Suppose [itex]\psi(\mathbf{x},t)[/itex] is an eigenfunction of [itex]H[/itex], [itex]\mathbf{L}^2[/itex], and [itex]L_z[/itex]. According to Weinberg, "since [itex]\mathbf{L}^2[/itex] acts only on angles, such a wavefunction must be proportional to a function only of angles, with a coefficient of proportionality [itex]R[/itex] that can depend only on [itex]r[/itex]. That is, for all r,
    [tex]\psi(\mathbf{x})=R(r)Y(\theta,\phi)."[/tex]

    He does not elaborate further on this, in my view, non-trivial statement. Can someone here provide a proof of his claim?
     
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  3. Sep 10, 2013 #2

    Nugatory

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    I'd be inclined to work on the intuition behind that claim instead of looking for a rigorous proof, because the general idea is useful in so many other problems.

    But first, which part of the statement are you asking about? The part about ##L^2## acting only on ##\theta## and ##\phi##, or the claim that this implies the ##R(r)Y(\theta,\phi)## form of the solution?
     
  4. Sep 10, 2013 #3
    I'm talking about the claim that [itex]\psi=R(r)Y(\theta,\phi)[/itex] given that [itex]\mathbf{L}^2[/itex] acts only on angles.
     
  5. Sep 10, 2013 #4

    dextercioby

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    This is trivial once you consider the uniparticle Hilbert space for the 'dummy' particle 'moving' aroung the CoM. Its Hilbert space is L^2(R^3) = L^2 (R) X L^2 (R) X L^2 (R) and after performing a unitary transformation of the whole space you have that L^2(R^3) ~ L^((0,infinity),dr) X L^2 (S^2,dOmega).

    In the position space the wavefunction of the 'dummy' particle would be then a product of a (sq integrable) function of r times a product of a function of spherical angles.
     
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